Questions
A uniform force of newton acts on a particle of mass 2 kg. Hence the particle is displaced from position meter to position meter. The work done by the force on the particle is
detailed solution
Correct option is C
Here,F→=3i^+j^N Intital position, r→1=2i^+k^m Final position, r→2=4i^+3j^−k^m Diplacement, r→=4i^+3j^−k^m−2i^+k^m =2i^+3j^−2k^m Work done, W=F→.r→=3i^+j^.2i^+3j^−2k^ =6+3=9JTalk to our academic expert!
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